Cap 20

20.1: a) [pic]
20.2: a) [pic]
20.3: a) [pic]
b) [pic]
c) [pic]
d) [pic]

20.4: a) [pic]
b)[pic]

20.5: a) [pic]

20.6: Solving [pic]
[pic]
[pic]
If the first equation is used (for instance, using a calculator without the [pic]function), note that the symbol “e” is theideal efficiency , not the base of natural logarithms.

20.7: a) [pic]
b) [pic]

20.8: a) From [pic]
b) [pic] an increase of 2%. If more figures are kept for the efficiencies, the difference is 1.4%.

20.9: a) [pic]
b)[pic]

20.10: [pic]

[pic]
20.11: a) [pic]or
[pic]

20.12: a) [pic]
[pic]
b) [pic]
c)[pic]
[pic]

20.13: a)[pic]
b) [pic]
c) [pic]

20.14: a) From Eq. (20.13), the rejected heat is [pic]
b) [pic]
c) From either Eq. (20.4) or Eq. (20.14), e=0.423=42.3%.

20.15: a) [pic]
[pic]
or [pic]
[pic][pic][pic]

20.16: a) From Eq. (20.13), [pic]b) The work per cycle is [pic] and [pic]keeping an extra figure.
c) [pic]

20.17: For all cases, [pic] a) Theheat is discarded at a higher temperature, and a refrigerator is required; [pic]
[pic] b) Again, the device is a refrigerator, and [pic] c) The device is an engine; the heat is taken form the hot reservoir, and the work done by the engine is [pic]
[pic]

20.18: For the smallest amount of electrical energy, use a Carnot cycle.
[pic]
Carnot cycle: [pic]

[pic][pic]
[pic]
20.19: The total work that must be done is

[pic]
[pic] Find [pic] so can calculate work [pic] done each cycle:
[pic]
[pic]
[pic]
The number of cycles required is [pic]

20.20: For a heat engine, [pic] [pic]This does not make use of the given value of [pic] [pic]whichgives the same result.

20.21: [pic]
[pic]

20.22: The claimed efficiency of the engine is [pic] While the most efficient engine that can operate between those temperatures has efficiency [pic]
The proposed engine would violate the second law of thermodynamics, and is not likely to find a market among the prudent.

20.23: a) Combining Eq. (20.14) and Eq. (20.15),

[pic]b) As [pic]and this is useless as a refrigerator. As [pic]engine does no work [pic]and a refrigerator that requires no energy input is very good indeed.

20.24: [pic]
[pic]
c) [pic] (If more figures are kept in the intermediate calculations, or if [pic]
[pic]

20.25: a) Heat flows out of the [pic] water into the ocean water and the [pic] water cools to [pic](the ocean warms, very, very slightly). Heat flow for an isolated system is always in this direction, from warmer objects into cooler objects, so this process is irreversible.

b) [pic]of water goes form [pic]
[pic]
This Q comes out of the 0.100 kg of water and goes into the ocean.
For the 0.100 kg of water,
[pic]
For the ocean the heat flow is [pic]and occurs at constant T:
[pic][pic]

20.26: (a) Irreversible because heat will not spontaneously flow out of 15 kg of water into a warm room to freeze the water.

(b) [pic]

[pic]

[pic]

[pic]

This result is consistent with the answer in (a) because [pic] for irreversible processes.

20.27: The final temperature will be

[pic]

and so theentropy change is
[pic]

20.28: For an isothermal expansion, [pic]

20.29: The entropy change is [pic] Thus,

[pic]

20.30: a) [pic] Note that this is the change of entropy of the water as it changes to steam. b) The magnitude of the entropy change is roughly five times the value found in Example 20.5. Water is less ordered (more random) than ice, but water is far less random than...